Source: Advanced Calculus I, University of Central Florida
Tags: u-substitution, u-sub, pattern recognition, integration techniques, reverse chain rule, antiderivatives, indefinite integrals, definite integrals, change of variable
Difficulty: Intermediate Prerequisites: Basic antiderivative rules (power rule, exponential, trigonometric), the chain rule for differentiation, properties of definite integrals.
u-Substitution is the single most important integration technique you will learn before integration by parts. It is, in essence, the chain rule run in reverse: where differentiation peels layers off a composition, u-sub rebuilds them. Nearly every integral you meet in this course and beyond will either yield directly to u-sub or require it as the first step inside a larger method. If you are comfortable with the chain rule, this topic is its mirror image. If the chain rule still feels shaky, revisit it first.
When an integrand contains a composite function and its inner function's derivative (or a constant multiple of it) also appears in the integrand, you can let u equal the inner function, replace dx with du, integrate the simpler expression, and substitute back. For definite integrals you can either convert the limits to u-values or back-substitute and use the original limits.
u-Substitution (change of variable)
A technique for evaluating integrals by introducing a new variable u = g(x) so that the integral simplifies. You replace every x-expression and dx with expressions in u and du. In simple terms, you are "undoing" a chain-rule derivative by identifying the inner function and its derivative inside the integrand.
Pattern recognition (reverse chain rule)
The ability to spot that an integrand is already in the form f'(g(x)) * g'(x) without needing to write out the full substitution. With practice, you evaluate these integrals by inspection. Think of it as doing u-sub mentally: you see the inner function, see its derivative sitting alongside, and write the antiderivative directly.
Inner function
The function you choose as u. A good choice is a function whose derivative (up to a constant factor) also appears in the integrand. In simple terms, it is the "inside" piece of a composition, the part nested inside a power, root, exponential, logarithm, or trig function.
Differential (du)
If u = g(x), then du = g'(x) dx. This relationship lets you swap dx for du. If the integrand contains g'(x) dx, that entire piece becomes du. If it contains a constant multiple of g'(x), you adjust with a constant factor.
Adjusting constants
When the integrand has g'(x) multiplied by a constant that does not match, you compensate by multiplying and dividing by the needed constant. For example, if du = 2x dx but you only see x dx, then x dx = du/2. You may only adjust by constants, never by functions of x.
Changing limits of integration
For a definite integral, once you set u = g(x), convert the original x-limits to u-limits: when x = a, u = g(a); when x = b, u = g(b). You then evaluate entirely in u and never back-substitute.
Look for a composite structure: something raised to a power, inside an exponential, under a root, inside a trig or log function.
Let u be the inner function of that composition.
Compute du and check whether g'(x) dx (or a constant multiple of it) accounts for whatever remains in the integrand.
If the leftover pieces are not just a constant, your u choice is probably wrong. Try a different candidate.
Identify u = g(x) and compute du = g'(x) dx.
Rewrite the entire integrand in terms of u and du. No x should remain.
Integrate with respect to u.
Substitute back: replace u with g(x).
Add + C.
Same first two steps as above.
Convert the limits: lower limit becomes g(a), upper limit becomes g(b).
Evaluate the antiderivative at the new limits. Do not convert back to x.
Power of an inner function: ∫ [g(x)]^n * g'(x) dx = [g(x)]^(n+1) / (n+1) + C, provided n ≠ -1.
Exponential of an inner function: ∫ e^(g(x)) * g'(x) dx = e^(g(x)) + C.
Logarithmic pattern: ∫ g'(x) / g(x) dx = ln|g(x)| + C.
Trig compositions: ∫ cos(g(x)) * g'(x) dx = sin(g(x)) + C, and similar for other trig functions.
Square root / fractional powers: treat √(g(x)) as [g(x)]^(1/2) and apply the power rule.
When given f''(x) and asked for f(x), you integrate twice. Each integration introduces one arbitrary constant, giving you Cx + D in the most general antiderivative. Watch the chain-rule constants: integrating cos(3x) gives sin(3x)/3, not sin(3x).
When given f'(x) and f(0) = some value, integrate f'(x) to get f(x) + C, then use the initial condition to solve for C. Only after finding C can you evaluate f at the requested point.
∫ cos(kx) dx = sin(kx)/k + C
∫ sin(kx) dx = -cos(kx)/k + C
∫ e^(kx) dx = e^(kx)/k + C
∫ 1/x dx = ln|x| + C
∫ sec²(x) dx = tan(x) + C (equivalently, d/dx[tan x] = sec²x = 1/cos²x)
(1 + e^(-x))² = 1 + 2e^(-x) + e^(-2x) (expand before integrating)
cos³(t) = cos(t)(1 - sin²(t)) (rewrite using the Pythagorean identity to set up u = sin t)
∫ 1/(1 + 9x²) dx = (1/3) arctan(3x) + C
u-Substitution is not a classroom-only technique. Any time a physical quantity is modelled as a rate of change of a composition (population growth with a time-dependent exponent, charge accumulation in a circuit with a nonlinear voltage function, heat transfer through a composite wall), the integral you need to evaluate uses exactly this method. Engineers and physicists use it so reflexively they rarely name it.
Students often forget to divide by the inner derivative's constant when integrating compositions like cos(3x). Integrating cos(3x) gives sin(3x)/3, not sin(3x). Always account for the chain rule constant.
Students sometimes try to "adjust" by a function of x rather than a constant. If after choosing u, the leftover factor is not a pure constant, u-sub alone will not work. You cannot pull x out of the integral if x depends on the variable of integration.
When changing limits on a definite integral, students occasionally forget to convert both limits or mix up which is which. If u = 1 + x² and x goes from 0 to 1, then u goes from 1 to 2, not from 0 to 1.
Students often drop the absolute value in ln|g(x)| when the denominator could be negative, particularly in problems with trig functions.
⚠️ u-Substitution appears on virtually every calculus exam. Expect at least one multiple-choice and one free-response question requiring it.
⚠️ Pattern recognition speed matters under time pressure. Practise until you can spot u and du within seconds.
⚠️ Definite integrals with u-sub are a common source of sign and limit errors. Double-check your converted limits before evaluating.
⚠️ Problems that combine u-sub with initial conditions (like Q3) test two skills at once and are a favourite for exam writers.
⚠️ Integrals of the form f'(x)/f(x) always give ln|f(x)| + C. Recognise this shape instantly.
True or false: If u = x² + 1, then du = 2x dx.
Fill in the blank: ∫ e^(5x) dx = ______.
True or false: When converting a definite integral to u-limits, you should substitute back to x before evaluating.
Fill in the blank: ∫ cos(4x) dx = ______.
True or false: You can compensate for a missing factor of x² by pulling it outside the integral sign.
Answers: 1. True. 2. e^(5x)/5 + C. 3. False, you evaluate directly with u-limits. 4. sin(4x)/4 + C. 5. False, you can only pull out constants.
This material connects directly to integration by parts (the next major technique), where u-sub often handles the remaining integral after applying the parts formula. It also underpins every substitution-based method you will meet later, including trigonometric substitution and partial fractions, both of which start by transforming the integral into a more workable form. The idea of reversing a differentiation rule carries through to solving separable differential equations, where you effectively u-sub both sides.
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